Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{y + 8}{y^3 + 6y^2 - 16y} \div \dfrac{3y - 12}{3y^3 - 27y^2 + 42y} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{y + 8}{y^3 + 6y^2 - 16y} \times \dfrac{3y^3 - 27y^2 + 42y}{3y - 12} $ First factor out any common factors. $n = \dfrac{y + 8}{y(y^2 + 6y - 16)} \times \dfrac{3y(y^2 - 9y + 14)}{3(y - 4)} $ Then factor the quadratic expressions. $n = \dfrac {y + 8} {y(y - 2)(y + 8)} \times \dfrac {3y(y - 2)(y - 7)} {3(y - 4)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {(y + 8) \times 3y(y - 2)(y - 7) } { y(y - 2)(y + 8) \times 3(y - 4)} $ $n = \dfrac {3y(y - 2)(y - 7)(y + 8)} {3y(y - 2)(y + 8)(y - 4)} $ Notice that $(y - 2)$ and $(y + 8)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {3y\cancel{(y - 2)}(y - 7)(y + 8)} {3y\cancel{(y - 2)}(y + 8)(y - 4)} $ We are dividing by $y - 2$ , so $y - 2 \neq 0$ Therefore, $y \neq 2$ $n = \dfrac {3y\cancel{(y - 2)}(y - 7)\cancel{(y + 8)}} {3y\cancel{(y - 2)}\cancel{(y + 8)}(y - 4)} $ We are dividing by $y + 8$ , so $y + 8 \neq 0$ Therefore, $y \neq -8$ $n = \dfrac {3y(y - 7)} {3y(y - 4)} $ $ n = \dfrac{y - 7}{y - 4}; y \neq 2; y \neq -8 $